Using the triangle inequalities to prove a bound in the complex plane

Question

Let $G$ be the interior of the circle in the complex plane given by $|z-(1+i)|=1$. Prove that if $z\in G$ then $\sqrt{5}-1<|z-3|<\sqrt{5}+1$.

Attempt so far: We know by the triangle inequality that $|a+b|\leq |a|+|b|$ and also $|a-b|\geq||a|-|b||$. Now, $G$ is a circle centred at $(1+i)$ with radius $1$. Using the inequality above, we have $|z-3|\leq |z|+3$, and so $|z-3|\leq 4$. For the lower bound, $|z-3|\geq ||z|-3|=|\sqrt{2}-3|$. This isn't the required inequality though. Any ideas?

Answer 1

If $\bigl|z-(1+i)\bigr|<1$, then

\begin{align}|z-3|&=\bigl|z-(1+i)+(1+i)-3\bigr|\\&\leqslant\bigl|z-(1+i)\bigr|+|-2+i|\\&<1+\sqrt5\end{align} and\begin{align}|z-3|&=\bigl|z-(1+i)+(1+i)-3\bigr|\\&\geqslant\bigl|\bigl|z-(1+i)\bigr|-|-2+i|\bigr|\\&=\bigl|\bigl|z-(1+i)\bigr|-\sqrt5\bigr|\\&>\sqrt5-1.\end{align}Besides, the bounds $\sqrt5+1$ and $\sqrt5-1$ are the best ones, as the picture below suggests.

Answer 2

Use $|z-3| \leqslant |z-(1+i)| + |(1+i)-3|$ instead.